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Math Help - Imaginary, arg problem

  1. #1
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    Imaginary, arg problem

    Hi I have this problem writting an expression in polar form, I have no issue finding the correct absolute value, but the argument turns out wrong.
    The expression is: \frac{(2+2i)(1+\sqrt{3})}{3i(\sqrt{12}-2i)}
    I wrote all the factors out in polar form, then added the arguments in the denominator and then subracted the arguments in the nominator, what do I do wrong?
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  2. #2
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    Quote Originally Posted by Henryt999 View Post
    Hi I have this problem writting an expression in polar form, I have no issue finding the correct absolute value, but the argument turns out wrong.
    The expression is: \frac{(2+2i)(1+\sqrt{3})}{3i(\sqrt{12}-2i)}
    I wrote all the factors out in polar form, then added the arguments in the denominator and then subracted the arguments in the nominator, what do I do wrong?
    Maybe convert it to a + ib form first.


    Also, should that numerator be (2 + 2i)(1 + \sqrt{3}) or (2 + 2i)(1 + \sqrt{3}\color{red}i\color{black})?
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    Quote Originally Posted by Henryt999 View Post
    Hi I have this problem writting an expression in polar form, I have no issue finding the correct absolute value, but the argument turns out wrong.
    The expression is: \frac{(2+2i)(1+\sqrt{3})}{3i(\sqrt{12}-2i)}
    I wrote all the factors out in polar form, then added the arguments in the denominator and then subracted the arguments in the nominator, what do I do wrong?

    Why don't you first simplify the mess? And I'm assuming you meant 1+\sqrt{3}i in the numerator:

    \frac{(2+2i)(1+\sqrt{3}i)}{3i(\sqrt{12}-2i)}=\frac{(1+i)(1+\sqrt{3}i)}{3i(\sqrt{3}-i)} =\frac{2e^{\pi i\slash 4}\cdot4e^{\pi i\slash 3}}{3e^{\pi i\slash 2}\cdot 4e^{-\pi i\slash 6}}

    Now simplify the above expression and you're on (note: I wrote down the principal argument in each case. In the second factor in the denominator you can write 4e^{5\pi i\slash 3} instead if you want all the arguments to be positive and in [0,2\pi]
    Last edited by tonio; March 7th 2010 at 05:50 AM. Reason: Correcting typo in the denominator
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    Hey thanks!

    First of all thanks to both of you for your help.
    You are correct there should be an i there where ProveIt said.
    Accutually I have to solve this in three different ways.
    By simplifying a+bi
    by rewritting it in polar form
    By using (as tonio did) Eulers approach.
    Also this has to be done without a calculator, (that probably means that the argument rational).
    Anyway the issue is that I donīt understand what I do wrong with the polar form thing: I just took the arguments in the denominator that gave me:
    \frac{\pi}{4}+\frac{\pi}{3} = \frac{7\pi}{12}

    then the arguments in the nominator: \frac{\pi}{2}+\frac{\pi}{6} = \frac{2\pi}{3}

    and then substracted that gave me \frac{7\pi}{12}-\frac{8\pi}{12} = -\frac{\pi}{12}

    but the correct answer is \frac{\pi}{4}
    not sure what Iīm doing wrong here...
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    \frac{(2+2i)(1+\sqrt{3}i)}{3i(\sqrt{12}-2i)}

     = \frac{2 + 2\sqrt{3}i + 2i - 2\sqrt{3}}{6 + 3\sqrt{12}i}

     = \frac{2[1 - \sqrt{3} + (1 + \sqrt{3})i]}{6 + 6\sqrt{3}i}

     =
    \frac{2[1 - \sqrt{3} + (1 + \sqrt{3})i]}{6(1 + \sqrt{3}i)}

     =
    \frac{1 - \sqrt{3} + (1 + \sqrt{3})i}{3(1 +  \sqrt{3}i)}

     =
    \frac{[1 - \sqrt{3} + (1 + \sqrt{3})i](1 - \sqrt{3}i)}{3(1 +  \sqrt{3}i)(1 - \sqrt{3}i)}


    You should be able to go from here (the bottom will become a rational number).
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    Quote Originally Posted by Henryt999 View Post
    First of all thanks to both of you for your help.
    You are correct there should be an i there where ProveIt said.
    Accutually I have to solve this in three different ways.
    By simplifying a+bi
    by rewritting it in polar form
    By using (as tonio did) Eulers approach.
    Also this has to be done without a calculator, (that probably means that the argument rational).
    Anyway the issue is that I donīt understand what I do wrong with the polar form thing: I just took the arguments in the denominator that gave me:
    \frac{\pi}{4}+\frac{\pi}{3} = \frac{7\pi}{12}

    then the arguments in the nominator: \frac{\pi}{2}+\frac{\pi}{6} = \frac{2\pi}{3}


    This is already a mess: in a fraction the upper part is called NUMERATOR and the lower one DENOMINATOR. Now, in the numerator you have z_1=1+i\,,\,\,z_2=1+\sqrt{3}\Longrightarrow \arg z_1=\arctan(1\slash 1)=\pi\slash 4 arg(z_2)=\arctan(\sqrt{3})=\pi\slash 3 , and in the denominator you have z_3=3i\,,\,\,z_4=\sqrt{3}-i\Longrightarrow \arg(z_3)=\pi\slash 2\,,\,\,\arg(z_4)=-\pi\slash 6.

    Now you do the maths with the powers and you'll get \pi\slash 4 , indeed.

    Tonio


    and then substracted that gave me \frac{7\pi}{12}-\frac{8\pi}{12} = -\frac{\pi}{12}

    but the correct answer is \frac{\pi}{4}
    not sure what Iīm doing wrong here...
    .
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